2 Answers
2

Since you ask only for a sufficient condition, let $l=2n$ be even and suppose that $\alpha$
is effective i.e.
a positive linear combination of fundamental classes of subvarieties $V_i$.
Let $[H]\in H^2(X)$ be the fundamental class of a hyperplane section. Then $\beta=\[H]^n\in H^l(X)$
is a class such that $\alpha\cup \beta\not=0$. To see this, note that after cupping with additional copies of $[H]$, we obtain the (positive weighted) sum of degrees of $V_i$.

Addendum (added in response to the edited question): The case where $l$ is odd is actually more interesting. Let me give an example
to show that the desired result can fail without additional hypotheses.
Choose a smooth projective variety $X'$ with $\dim X'=n>1$ and $H^l(X')\not=0$ with $l$ odd. Now blow up a smooth codimension $k+1>1$ subvariety $V$ to get $X$. Then $H^l(X')\subset H^l(X)$, so it's also nonzero.
Let $E$ be the exceptional divisor. This is a $\mathbb{P}^{k}$ bundle over $V$.
Let $\alpha=[\mathbb{P}^k]$ (one of the fibres). This is a nonzero class, but $\alpha\cup:H^l(X)\to H^{l+2k}(X)$ is zero, because it factors through restriction to $\mathbb{P}^{k}$. Note that $\dim H^l$ and $\dim H^{l+2k}$ can be arbitrarily large.

Thank you very much for your answer. The point here is that I do not have much control on $\alpha$, and so I do not knot how to relate it to the hyperplane (or kahler) class. Is there some trick (or philosophy) to circumvent this problem?
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calcMay 4 '12 at 17:53

Do you know if it is a top Chern class of some vector bundle?
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Vladimir BaranovskyMay 7 '12 at 4:09

Yes, $\alpha$ is the top Chern class of a vector bundle. Does this help?
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calcMay 9 '12 at 6:27